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Question Number 22874 by selestian last updated on 23/Oct/17

Answered by $@ty@m last updated on 23/Oct/17

Solutionof 9. S_1 =((n(n+1))/2) S_2 =((n(n+1)(n+2))/6) S_3 ={((n(n+1))/2)}^2 ∴((S_3 (1+8S_1 ))/S_2 ^2 )=((n^2 (n+1)^2 .{1+8×((n(n+1))/2)})/(4×((n^2 ×(n+1)^2 ×(2n+1)^2 )/(36)))) =(({1+4n(n+1)}×9)/((2n+1)^2 )) =((9(4n^2 +4n+1))/((2n+1)^2 )) =9

$${Solutionof}\:\mathrm{9}. \\ $$$${S}_{\mathrm{1}} =\frac{{n}\left({n}+\mathrm{1}\right)}{\mathrm{2}} \\ $$$${S}_{\mathrm{2}} =\frac{{n}\left({n}+\mathrm{1}\right)\left({n}+\mathrm{2}\right)}{\mathrm{6}} \\ $$$${S}_{\mathrm{3}} =\left\{\frac{{n}\left({n}+\mathrm{1}\right)}{\mathrm{2}}\right\}^{\mathrm{2}} \\ $$$$\therefore\frac{{S}_{\mathrm{3}} \left(\mathrm{1}+\mathrm{8}{S}_{\mathrm{1}} \right)}{{S}_{\mathrm{2}} ^{\mathrm{2}} }=\frac{{n}^{\mathrm{2}} \left({n}+\mathrm{1}\right)^{\mathrm{2}} .\left\{\mathrm{1}+\mathrm{8}×\frac{{n}\left({n}+\mathrm{1}\right)}{\mathrm{2}}\right\}}{\mathrm{4}×\frac{{n}^{\mathrm{2}} ×\left({n}+\mathrm{1}\right)^{\mathrm{2}} ×\left(\mathrm{2}{n}+\mathrm{1}\right)^{\mathrm{2}} }{\mathrm{36}}} \\ $$$$=\frac{\left\{\mathrm{1}+\mathrm{4}{n}\left({n}+\mathrm{1}\right)\right\}×\mathrm{9}}{\left(\mathrm{2}{n}+\mathrm{1}\right)^{\mathrm{2}} } \\ $$$$=\frac{\mathrm{9}\left(\mathrm{4}{n}^{\mathrm{2}} +\mathrm{4}{n}+\mathrm{1}\right)}{\left(\mathrm{2}{n}+\mathrm{1}\right)^{\mathrm{2}} } \\ $$$$=\mathrm{9} \\ $$

Answered by $@ty@m last updated on 23/Oct/17

Solution of 10. a_1 , a_2 , a_3 ... a_n are in HP ⇒(1/a_1 ), (1/a_2 ),...(1/a_n ) are in AP ⇒a_2 =((2a_1 a_3 )/(a_1 +a_3 )) ⇒a_1 a_2 +a_2 a_3 =2a_1 a_3 −−−(1) Similarly a_3 a_4 +a_4 a_5 =2a_3 a_5 −−−(2) Now (1/a_3 )−(1/a_1 )=(1/a_5 )−(1/a_3 ) ⇒(2/a_3 )=((a_1 +a_5 )/(a_1 a_5 )) ⇒2a_1 a_5 =a_1 a_3 +a_3 a_5 ×2 ⇒4a_1 a_5 =2a_1 a_3 +2a_3 a_5 ⇒a_1 a_2 +a_2 a_3 +a_3 a_4 +a_4 a_5 =4a_1 a_5 −−(3) {using (1) & (2) Looking at the pattern of (1) and (3) we can easily conclude that a_1 a_2 +a_2 a_3 +a_3 a_4 +.....+a_(n−1) a_n =(n−1)a_1 a_n ⇒k=n−1 Ans.

$${Solution}\:{of}\:\mathrm{10}. \\ $$$${a}_{\mathrm{1}} ,\:{a}_{\mathrm{2}} \:,\:{a}_{\mathrm{3}} \:...\:{a}_{{n}} \:\:{are}\:{in}\:{HP} \\ $$$$\Rightarrow\frac{\mathrm{1}}{{a}_{\mathrm{1}} },\:\frac{\mathrm{1}}{{a}_{\mathrm{2}} },...\frac{\mathrm{1}}{{a}_{{n}} }\:\:{are}\:{in}\:{AP} \\ $$$$\Rightarrow{a}_{\mathrm{2}} =\frac{\mathrm{2}{a}_{\mathrm{1}} {a}_{\mathrm{3}} }{{a}_{\mathrm{1}} +{a}_{\mathrm{3}} } \\ $$$$\Rightarrow{a}_{\mathrm{1}} {a}_{\mathrm{2}} +{a}_{\mathrm{2}} {a}_{\mathrm{3}} =\mathrm{2}{a}_{\mathrm{1}} {a}_{\mathrm{3}} \:\:−−−\left(\mathrm{1}\right) \\ $$$${Similarly} \\ $$$${a}_{\mathrm{3}} {a}_{\mathrm{4}} +{a}_{\mathrm{4}} {a}_{\mathrm{5}} =\mathrm{2}{a}_{\mathrm{3}} {a}_{\mathrm{5}} \:\:−−−\left(\mathrm{2}\right) \\ $$$${Now}\:\frac{\mathrm{1}}{{a}_{\mathrm{3}} }−\frac{\mathrm{1}}{{a}_{\mathrm{1}} }=\frac{\mathrm{1}}{{a}_{\mathrm{5}} }−\frac{\mathrm{1}}{{a}_{\mathrm{3}} } \\ $$$$\Rightarrow\frac{\mathrm{2}}{{a}_{\mathrm{3}} }=\frac{{a}_{\mathrm{1}} +{a}_{\mathrm{5}} }{{a}_{\mathrm{1}} {a}_{\mathrm{5}} } \\ $$$$\Rightarrow\mathrm{2}{a}_{\mathrm{1}} {a}_{\mathrm{5}} ={a}_{\mathrm{1}} {a}_{\mathrm{3}} +{a}_{\mathrm{3}} {a}_{\mathrm{5}} \:\:\:\:\:×\mathrm{2} \\ $$$$\Rightarrow\mathrm{4}{a}_{\mathrm{1}} {a}_{\mathrm{5}} =\mathrm{2}{a}_{\mathrm{1}} {a}_{\mathrm{3}} +\mathrm{2}{a}_{\mathrm{3}} {a}_{\mathrm{5}} \\ $$$$\Rightarrow{a}_{\mathrm{1}} {a}_{\mathrm{2}} +{a}_{\mathrm{2}} {a}_{\mathrm{3}} +{a}_{\mathrm{3}} {a}_{\mathrm{4}} +{a}_{\mathrm{4}} {a}_{\mathrm{5}} =\mathrm{4}{a}_{\mathrm{1}} {a}_{\mathrm{5}} \:\:−−\left(\mathrm{3}\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\left\{{using}\:\left(\mathrm{1}\right)\:\&\:\left(\mathrm{2}\right)\right. \\ $$$${Looking}\:{at}\:{the}\:{pattern}\:{of}\:\left(\mathrm{1}\right)\:{and}\:\left(\mathrm{3}\right) \\ $$$${we}\:{can}\:{easily}\:{conclude}\:{that} \\ $$$${a}_{\mathrm{1}} {a}_{\mathrm{2}} +{a}_{\mathrm{2}} {a}_{\mathrm{3}} +{a}_{\mathrm{3}} {a}_{\mathrm{4}} +.....+{a}_{{n}−\mathrm{1}} {a}_{{n}} =\left({n}−\mathrm{1}\right){a}_{\mathrm{1}} {a}_{{n}} \\ $$$$\Rightarrow{k}={n}−\mathrm{1}\:{Ans}. \\ $$

Answered by math solver last updated on 23/Oct/17

Commented by math solver last updated on 23/Oct/17

alternative solution of q.10.... if you find difficulty in understanding the pattern (ans. given by satyam bro) :)

$${alternative}\:{solution}\:{of}\:{q}.\mathrm{10}.... \\ $$$${if}\:{you}\:{find}\:{difficulty}\:{in}\:{understanding} \\ $$$$\left.{the}\:{pattern}\:\left({ans}.\:{given}\:{by}\:{satyam}\:{bro}\right)\::\right) \\ $$

Commented by $@ty@m last updated on 23/Oct/17

Nice.. It′s the systematic solution.

$${Nice}.. \\ $$$${It}'{s}\:{the}\:{systematic}\:{solution}. \\ $$

Commented by math solver last updated on 23/Oct/17

thanks ! ;)

$$\left.{thanks}\:!\:;\right) \\ $$

Commented by selestian last updated on 24/Oct/17

thAnks sir systematic is easy to understand

$${thAnks}\:{sir}\:{systematic}\:{is}\:{easy}\:{to} \\ $$$${understand} \\ $$

Answered by Rasheed.Sindhi last updated on 25/Oct/17

Q#9 Easy solution! ((S_3 (1+8S_1 ))/S_2 ^( 2) ) is constant for all n∈N n=1⇒S_1 =S_2 =S_3 =1 ((S_3 (1+8S_1 ))/S_2 ^( 2) )=((1(1+8×1))/1^2 )=9 You need to evaluate only for one value of n(It′s sufficient) and simple case is n=1.

$$\mathcal{Q}#\mathrm{9} \\ $$$$\mathcal{E}{asy}\:{solution}! \\ $$$$\frac{\mathrm{S}_{\mathrm{3}} \left(\mathrm{1}+\mathrm{8S}_{\mathrm{1}} \right)}{\mathrm{S}_{\mathrm{2}} ^{\:\mathrm{2}} }\:\mathrm{is}\:\mathrm{constant}\:\mathrm{for}\:\mathrm{all}\:\mathrm{n}\in\mathbb{N} \\ $$$$\mathrm{n}=\mathrm{1}\Rightarrow\mathrm{S}_{\mathrm{1}} =\mathrm{S}_{\mathrm{2}} =\mathrm{S}_{\mathrm{3}} =\mathrm{1} \\ $$$$\frac{\mathrm{S}_{\mathrm{3}} \left(\mathrm{1}+\mathrm{8S}_{\mathrm{1}} \right)}{\mathrm{S}_{\mathrm{2}} ^{\:\mathrm{2}} }=\frac{\mathrm{1}\left(\mathrm{1}+\mathrm{8}×\mathrm{1}\right)}{\mathrm{1}^{\mathrm{2}} }=\mathrm{9} \\ $$$$\mathrm{You}\:\mathrm{need}\:\mathrm{to}\:\mathrm{evaluate}\:\mathrm{only}\:\mathrm{for} \\ $$$$\mathrm{one}\:\mathrm{value}\:\mathrm{of}\:\mathrm{n}\left(\mathrm{It}'\mathrm{s}\:\mathrm{sufficient}\right) \\ $$$$\mathrm{and}\:\mathrm{simple}\:\mathrm{case}\:\mathrm{is}\:\mathrm{n}=\mathrm{1}. \\ $$

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